I have a function $g(x)=f(x)e^{-x}$ and i want to consider the following integral: $\int_{0}^{\infty}g(x)dx$.
Since $f(x)$ is a complicated, but monotonic decreasing, function in the interval $[0,\infty[$ and it is multiplied by $e^{-x}$, the integrand $g(x)$ is a very rapidly decreasing function.
If I consider $T_{n}(x)$ a Taylor expansion of the function $f(x)$ around $x=0$ truncated at a certain order $n$, the integral $\int_{0}^{\infty}T_{n}(x)e^{-x}dx$ converges for every $n$.
What is the error committed in approximating $\int_{0}^{\infty}g(x)dx$ with $\int_{0}^{\infty}T_{n}(x)e^{-x}dx$?
and what would it be if $g(x)=f(x)\cdot x(e^{-x})$?
thanks in advance
Hint
If you have the Taylor expansion $$f(x)=\sum_{i=0}^n a_i x^i$$, you then have $$\int_0^\infty f(x)e^{-x}dx=\sum_{i=0}^n a_i i!$$ since $\int_0^\infty x^n e^{-x}=n!$.
I am sure that you can take from here.